Imagine there's a strong laser beam, not just an attenuated stream of single photons, entering a balanced Mach-Zehnder interferometer.

One-photon picture: Each photon interferes with itself on the second beam splitter. As a result, all photons leave out of the same output.

More realistic picture: If the intensity of the laser is strong enough, a lot of photons will enter the interferometer within the coherence time and will be therefore indistinguishable (at least by their arrival time).

How does this fact change the quantum mechanical description of the state? Shouldn’t there be a two-photon interference at the second beam splitter in this case? How do you describe the evolution of the state throughout the interferometer mathematically: two-photon interference of photons that are themselves in the same superposition of two paths?

  • $\begingroup$ to be clear, are you asking how you calculate the evolution through a MZ for coherent inputs and for single-photon inputs? What do you mean "how does this fact change the quantum mechanical description" then? It changes the description because you are using different inputs. $\endgroup$
    – glS
    Aug 30, 2019 at 13:55
  • $\begingroup$ @glS For a coherent state input. Any pointers for an easy intro into interference at beam splitters for coherent states? And why it is different from a single-photon interference? $\endgroup$
    – triclope
    Aug 31, 2019 at 0:41

2 Answers 2


It depends on the quantum state that entered the interferometer. In general the sate is described by a coherent state. To see what happens quantum mechanically one can apply the unitary operator for a beam splitter on the coherent state (with a tensor product with a vacuum state, because we assume nothing enters in the other port). What one would find is a tensor product of two coherent states in the respective path inside the interferometer, each with a slightly reduced amplitude. Inside the interferometer the two path may be slightly different, leading to a relative phase shift, which is easily represented in therms of the complex parameters of the two coherent states. Finally we send them through another beam splitter. At the two output ports one now obtain two coherent states whose amplitudes are respectively the sum and difference of the two input coherent states. The sum and difference are responsible for the interference we observe at the respective output ports.

In conclusion, we do not see quantum interference. Only optical interference, even when we do the calculation in quantum mechanics.


A MZ interferometer is very carefully setup, you can think of the path distance in 1 arm to be close to a perfect distance of n wavelengths of light and the other arm to be a n+1/2 wavelengths of light. Single photons entering the MZ interferometer will have the highest probability of transmission in the n arm and almost 0 probability in the n+1/2 arm.

Feynman's concept of the path integral and the photon wave function says the light will take the shortest path that is n multiples. If many photons are fired the result would be almost all photons in the n arm.


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